Logique et Démonstration : Éléments Essentiels

p
p
p q r, . . .
p q
p¬p p
p¬p
p q p q p q
p q p q p ∧q p ∨q
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p p
¬p
p q
p∧q p ∨q
p q
p⊕q
p q p ⊕q
p q q p (¬p)∨q
q p p ⇒q p q
p⇒q p q
p q p ⇒q
p⇒q p q p
q p q
q⇒p p ⇒q
¬q⇒ ¬p p ⇒q
p ABC
q ABC p ⇒q ABC
ABC
p q p q
(p⇒q)∧(q⇒p)p⇔q p q
p⇔q p, q
p q p ⇔q
p⇔q p q
p q p q
p, q p ⇔q
ABC ABC
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p
p∨(¬p)
p∧(¬p)
p, q r
(p∧p)⇔p
(p∨p)⇔p
(p∧q)⇔(q∧p)
(p∨q)⇔(q∨p)
[(p∧q)∧r]⇔[p∧(q∧r)]
[(p∨q)∨r]⇔[p∨(q∨r)]
[p∧(q∨r)] ⇔[(p∧q)∨(p∧r)]
[p∨(q∧r)] ⇔[(p∨q)∧(p∨r)]
[¬(¬p)] ⇔p
[¬(p∧q)] ⇔[(¬p)∨(¬q)]
[¬(p∨q)] ⇔[(¬p)∧(¬q)]
(p⇒q)⇔[(¬q)⇒(¬p)]
[(p⇒q)∧(q⇒r)] ⇒(p⇒r)
[(p⇒q)∧(q⇒r)∧(r⇒p)] ⇔[(p⇔q)∧(q⇔r)∧(p⇔r)]
[p∧(p⇒q)] ⇒q
p p
p⇔q p q
p q
(p⇒q)⇔[(¬p)∨q]
p[(p∨q)∧
(¬((¬p)∧q)]
E
E
E, F, . . . x, y, . . .
x, y E
x=y x y E
x6=y[¬(x=y)]
x∈E x E
x /∈E[¬(x∈E)]
∅
x{x}
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∀∃ ∃!
E={n∈N/n > 2}x E x
∀x∈E, x > 2
∀x∈E, P (x)P(x)x
∀x∈E, x > 2∃x∈E:x≤2
x > 2x
>2x
P(x)x > 4P(5) 5 >4P(3)
3>4
P(x, , y, z)z=x+y P (1,3,4) 4 = 1 + 3
P(x)x
E P Q
[¬(∀x∈E, P (x))] ⇔ ∃x∈E, ¬(P(x))
[¬(∃x∈E, P (x))] ⇔ ∀x∈E, ¬(P(x))
[∀x∈E, (P(X)∧Q(x)] ⇔[(∀x∈E, P (x)) ∧(∀x∈E, Q(x))]
[∃x∈E, (P(X)∨Q(x))] ⇔[(∃x∈E, P (x)) ∨(∃x∈E, Q(x))]
(un)n∈N
(un)⇔(∃l∈R) (∀ > 0),(∃N∈N):(∀n∈N)
n≥N⇒ |un−l| ≤ 
(un)⇔(∀l∈R),(∃ > 0) : (∀N∈N)(∃n∈N)
n≥N|un−l|> 
∀x∈R?,∃y∈R?:xy = 1 ∃y∈R?:∀x∈R?, xy = 1
P(x)x
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E P (x)
Q(x)x x
p q r
p⇒q p
q p
p⇒r1, r1⇒r2, . . . , rn⇒q r1, . . . , rnq
m n mn
n m k, h ∈Zm= 2h+1 n= 2k+1
mn = (2h+ 1)(2k+ 1) = 2(2hk +h+k)+1 mn
r
p∨q p ⇒r q ⇒r
n∈Nn(n+1)
2∈Nn∈Nn n
nn
2∈Nn(n+1) 2=n 2(n+ 1) ∈N
n n + 1 n+1
2∈Nn(n+1) 2=nn+1 2∈N
p⇒q¬q⇒ ¬p
p⇒q¬q⇒ ¬p
m, n ∈Na=mn m ≤√a n ≤√a
m > √a n > √a mn > √a√a=a a 6=mn
E P
E P
E={n∈N/n > 1} ∀n∈E, n −1∈E
k= 2 ∈E k −1=1/∈E
[(∀x∈E, P (x)) ∨(∀x∈E, Q(x))]
[∀x∈E, (P(x)∨Q(x)]
[(∃x∈E, P (x)) ∧(∃x∈E, Q(x))]
[∃x∈E, (P(x)∧Q(x)]
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